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3 years ago

5

Following are the different layers of the sun’s atmosphere. Rank them based on the order in which a probe would encounter them w

hen traveling from earth to the sun’s surface

1 answer:

blsea [12.9K]3 years ago

6 0

Answer:

Going from earth to the sun a probe would encounter the next layers in order:

Corona
Transition Region
Chromosphere
Photosphere
Convection Zone
Radiative Zone
Core

A brief description of them:

Corona is the outermost layer and it cannot be seen with the naked eye, is starts at about 2100 km from the surface of the sun and it has no limit defined.

Transition Region is between the corona and the chromosphere, it has an extension of about 100km

The chromosphere is between 400 km from the surface of the sun to 2100 km. In this layer the further you get away from the sun it gets hotter.

The photosphere is the surface of the sun, the part that we can see, and extends from the surface to 400km.

The convection zone is where convection happens, hot gas rises, cools and rises again.

Radiative Zone is where the photons try to rise to move to higher layers.

The core of the Sun is where nuclear fusion occurs due to the very high temperatures.

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A total of 25.6 kJ of heat energy is added to a 5.46 L sample of helium at 0.991 atm. The gas is allowed to expand against a fix

bagirrra123 [75]

Answer:

(a) W = 1329.5 J = 1.33 KJ

(b) ΔU = 24.27 KJ

Explanation:

(a)

Work done by the gas can be found by the following formula:

$W = P\Delta V$

where,

W = Work = ?

P = constant pressure = (0.991 atm)( $\frac{101325\ Pa}{1\ atm}$ ) = 100413 Pa

ΔV = Change in Volume = 18.7 L - 5.46 L = (13.24 L)( $\frac{0.001\ m^3}{1\ L}$ ) = 0.01324 m³

Therefore,

W = (100413 Pa)(0.01324 m³)

<u>W = 1329.5 J = 1.33 KJ</u>

<u></u>

(b)

Using the first law of thermodynamics:

ΔU = ΔQ - W (negative W for the work done by the system)

where,

ΔU = change in internal energy of the gas = ?

ΔQ = heat added to the system = 25.6 KJ

Therefore,

ΔU = 25.6 KJ - 1.33 KJ

<u>ΔU = 24.27 KJ</u>

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3 years ago

Which is the most accurate name for the covalent compound P2O3?

dexar [7]

P2o3 is <span>diphosphorus trioxide.
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3 years ago

Read 2 more answers

The conservation of momentum is most closely related to

BartSMP [9]

Answer:

<h2>B) Newton's 2nd law</h2>

Explanation:

<h2>From; force= mass × acceleration </h2><h2> f= m×a </h2><h2>where a(acceleration)= velocity/time</h2><h3> force = mv/t</h3><h3>But momentum(p) = Mass × velocity </h3><h2>hence force =p/t </h2><h3>that is Momentum = force × time ( Newton's 2nd law)</h3>

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1 year ago

Help with this physics task pls

cupoosta [38]

Answer:

Answers can be seen below

Explanation:

First we must explain the essential when we clear equations, and that is that if the term we need to clear is accompanied by other terms that are being added up, then those terms go to the other side of the equation to subtract if those terms are subtracting, then they go to the other side to add, if those terms are found multiplying then they go to the other side of the equation to divide and if those other terms are found dividing then they go to the other side of the equation to multiply.

(Primero debemos explicar lo esencial cuando despejamos ecuaciones, y es que si el término que necesitamos despejar va acompañado de otros términos que se están sumando, entonces esos términos van al otro lado de la ecuación para restar si esos términos están restando, luego van al otro lado para sumar, si esos términos se encuentran multiplicando luego van al otro lado de la ecuación a dividir, y si esos términos se encuentran dividiendo, pasan al otro lado de la ecuación a multiplicar.)

1 )

$t=\frac{v}{a}$ ; $d=s*(t-t_{0} )$

2)

$k=\frac{2*U}{x^{2} }$ ; $T_{2}=\frac{P_{2}*V_{2}*T_{1} }{P_{1}*V_{1} } \\$

3)

$L=\frac{F}{\pi*r*P}$ ; $d=\frac{w}{F*cos(o)}$

4)

$t^{2}=\frac{2*x}{g} ; V_{2}=\frac{A_{1}*V_{1} }{A_{2} } \\$

5)

$h=\frac{V}{\pi *r^{2} } ; r=\frac{t}{F*sin(o)}$

6)

$h=\frac{m}{(1/2)*\pi *r^{2} } ; h_{2}=\frac{F_{2}*(1/2)*b_{1} *h_{1} }{F_{1}*(1/2)*b_{2}*h_{2} }$

7)

$b=\frac{mg-ma}{v}; m=\frac{F+kx}{g*cos(o)}$

8)

$a=\frac{v-v_{o} }{t} ; u=\frac{m_{1}+m_{2} }{M}$

9)

$v_{o}=\frac{x-\frac{1}{2}*a*t^{2} }{t} ; F=\frac{W+uNd}{d*cos(o)}$

10)

$h=\frac{E-\frac{1}{2}*m*v^{2} }{mg} ; v_{2} ^{2} = \frac{Dk-\frac{1}{2} m*v_{1}^{2} }{\frac{1}{2}m }$

11)

$N=\frac{mg*sin(o)-F}{u} ; x^{2}=\frac{W+\frac{1}{2}k*x_{1}^{2} }{\frac{1}{2}*k }$

12)

$x=x_{o} +\frac{v^{2-v_{o}^{2} } }{2a} ; m=\frac{P*A-F_{1}-F_{2} }{g}$

13)

$x_{o} = x-\frac{F}{k} ; u=\frac{cos(o)-\frac{a}{g} }{sin(o)}$

14)

$t=\frac{d}{v} +t_{o}$ ; $t_{o} = t-(\frac{v-v_{o} }{a} )$

15)

$F_{2}=\frac{W-F_{1} *d}{d}+F_{3} ; v_{2}^{2}=v_{1}^{2}+\frac{2*Dk}{m}$

16)

$y_{1}=y-\frac{u}{mg} ; x^{2} = \frac{2W}{k}+x_{o} ^{2}$

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3 years ago

The type of force measured by grocery store spring scale is

alexandr402 [8]

Answer:

Weight

Explanation:

The spring balance is used to measure weight of an object.

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